Download Project Scheduling: PERT/CPM

Project Scheduling: PERT/CPM
Project Scheduling with Known Activity Times (as in exercises 1, 2, 3 and 5 in the
handout) and considering Time-Cost Trade-Offs (as in exercises 4 and 6 in the
handout)
This is Critical Path Method
.
Project Scheduling with Uncertain Activity Times (as in exercises 7, 8 and 9 in the
handout).
This is Program Evaluation Research Technique
Acknowledgement: I have adapted the format and some ideas from the PowerPoint slides of Dr. C.
Lightner of Fayetteville State University, but most of the mathematical solutions and method I have
used are quite different and extend his work.
1
Introduction to Project Management
!   Project Scheduling or project management is used to schedule, manage and control
projects which can be analyzed into various semi-independent activities or tasks.
!   Example: Building a New Home
When building a home individual subcontractors are hired to:
―  Grade and prepare the land
―  Build the foundation
―  Frame up the home
―  Insulate the home
―  Wire (Electricity, Cable, Telephone lines) the home
―  Drywall
―  Paint (inside)
―  Put vinyl siding on home
―  Install Carpet
―  Landscape
―  Lay Concrete
2
PERT/CPM
!
!
PERT
–  Program Evaluation and Review Technique
–  Developed by U.S. Navy for Polaris missile project
–  Developed to handle uncertain activity times
CPM
–  Critical Path Method
–  Developed by Du Pont & Remington Rand
–  Developed for industrial projects for which activity times are known
!   There are project management software packages that can perform both.
3
PERT/CPM
!   PERT and CPM have been used to plan, schedule, and control a wide variety of
projects:
–  R&D of new products and processes
–  Construction of buildings and highways
–  Maintenance of large and complex equipment
–  Design and installation of management systems
–  Organizing transportation projects
–  Deployment and/or relocation of forces
–  Design of computer systems
4
PERT/CPM
!   PERT/CPM is used to plan the scheduling and optimal staffing of individual
activities that make up a project.
!   Projects may have as many as several thousand activities and may have to be
broken up into simpler sub-projects.
!   Usually some activities depend on the completion of other activities before they
can be started.
!   So we need to start with the Prerequisites Task Set giving the order of
precedencies, along with durations for each task, or activity
5
PERT/CPM
!   Project managers rely on PERT/CPM to help them answer questions such as:
–  What is the total time to complete the project?
–  What are the scheduled start and finish dates for each specific activity?
–  Which activities are critical and must be completed exactly as scheduled to
keep the project on schedule?
–  How long can noncritical activities be delayed before they cause an increase in
the project completion time?
–  If activity times are uncertain (i.e., they are random variables), what is the
probability that the project will be finished in any given time-frame?
–  If there is a cost associated with “crashing” each activity, what is the extra cost
of completing the project at a time earlier than the normal time?
6
Project Network
!   A project network can be constructed to model the precedence of the activities. It
consists of nodes and arrows between the nodes.
!   Each activity is represented by an arrow, which starts at a node and ends at a
different node.
!   The nodes of the network represent events comprising the completion of a subset
of activities, so that subsequent activities can occur.
!   The start node has only arrows issuing from it; the finish node has only arrows
entering it. All other nodes have arrows both entering and leaving.
! Immediate predecessor(s) is (are) activities that must be completed immediately
before a given activity can begin. The arrows entering a node represent the
immediate predecessors of those exiting the node.
!   The network is a representation of the precedence relationships of the activities.
7
The Critical Path and Activity Slack
!   A path through a network is one of the routes following the arrows (arcs) from the
start node to the finish node.
!   The length of a path is the sum of the (estimated) durations of the activities on the
path.
!   The (estimated) project duration or project completion time equals the length of
the longest path through the project network.
!   This longest path is called the critical path. (If more than one path tie for the
longest, they are all critical paths.)
! Activity Slack is the amount of time that noncritical activities can be delayed
without increasing the project completion time.
!   The critical path will always consist of activities with zero slack.
8
Constructing Project Networks
!   The Mohawk Discount Store is designing a management training program for
individuals at its corporate headquarters. The company wants to design a
program so that trainees can complete it as quickly as possible. Important
precedence relationships must be maintained between assignments or activities
in the program. For example, a trainee cannot serve as an assistant to the store
manager until the employee has obtained experience in the credit department and
at least one sales department. The following activities are the assignments that
must be completed by each program trainee. Construct the project network for
this problem. (Anderson, et. Al, Chapter 10, problem 1).
!   Try this yourself before looking at the next slide!
Activity
Activities A – H represent actual tasks.
A
B
C
D
Immediate
Predecessor
---
---
A
A, B
E
F
G
H
A,B
C
D,F
E,G
9
Mohawk Project Network
A
Start
C
D
B
F
G
H
Finish
E
10
Project Network Discussion
!   Project networks are not unique. A project network is considered valid provided
all precedence relationships are preserved.
!   If it is convenient, you can introduce an activity with zero duration, a so-called
dummy activity, so as to preserve precedence relationships. Sometimes you can
simplify a network by re-drawing it with the arrows in different positions (but with
the same precedence relations). Try to minimize the use of dummy activities.
!   ONLY ACTIVITIES THAT ARE NOT IMMEDIATE PREDECESSORS TO ANY
OTHER NETWORK ACTIVITIES MAY HAVE A LINK DIRECTLY TO THE FINISH
NODE.
!   No dangling activities (i.e., with no beginning or end node) and no loops are
allowed in the diagram.
11
Overall Procedure for solving a Project Network
1.  Determine the sequence of activities.
2.  Construct the network or precedence diagram. Make sure there are no
“dangling” activities or loops: all activities lead towards the end of the
project.
3.  Starting from the left, compute the Early Start (ES) time at each node. This
is the FORWARD PASS.
4.  Starting from the right, compute the Late Finish (LF) time at each node. This
is the BACKWARD PASS.
5.  Find the slack(s) for each activity.
6.  Identify the Critical Path.
In the following slides will elaborate on steps 3-6.
12
Project Management Notation: Nodes
ET
LT
tij
The earliest time all the activities entering a node can be said to be
completed. This will be computed on a forward pass through the
network, starting at the start node.
The latest time the node at the end of the activity can be completed and
not delay the project. This will be computed on a backward pass
through the network, starting at the finish node.
Duration of activity (i,j) (shorthand for the activity going from node i to
node j).
Node i
i, LTi ; ETi
Node j
Activity (i,j)
j, LTj ; ETj
duration tij
13
Example: Horizon Cable
!   Horizon is about to expand its cable TV offerings in Smalltown, adding the
Cooking Network and others. The activities in the table below are needed to
complete the service expansion.
Activity
Description
Immediate
Predecessors
Duration (weeks)
A
Choose stations
-
2
B
Get town council
approval
A
4
C
Order converters
B
3
D
Install new dish to
receive new stations
B
2
E
Install converters
C, D
10
F
Change billing
system
B
4
14
Example: Horizon Cable
2
B
Start
A 1
We put in the Start node on the left and see that activity A is
the only one with no predecessors. This is the first arrow,
leading to node 1. Then we see that the only activity with A as
predecessor is B, so we attach an arrow representing B to
Node 1, with Node 2 at its head.
15
Example: Horizon Cable
F
2
B
A 1
Start
Immediate Predecessors
Activity
A
-
B
A
C
B
D
B
E
C, D
F
B
C
D
3
Now from the table of
precedencies, we see that
activities C, D and F are
those that have B as an
immediate predecessor, so
we attach an arrow for each
with its base at node 2.
Activities C and D both end
in a single node # 3– to the
right of node 2. This is
because a single activity, E,
depends on the completion
of both of them together.
F has no ending node just yet: no activity has F as a
predecessor, so it will go to the End node.
16
Example: Horizon Cable
F
2
C
B
A 1
Start
D
Finish
E
3
Immediate Predecessors
Activity
A
-
B
A
C
B
D
B
E
C, D
F
B
Since no activities depend for their start on E and F,
both of these end in the Finish node and we are done
drawing the precedence relationship as a network.
It’s considered best to avoid having arrows crossing
each other. While this is not exactly wrong, neither is it particularly aesthetic,
so we try to avoid such crossings.
17
Earliest Node, Start and Finish Times
!   Step 3: Make a forward pass through the network as follows: For each node i
beginning at the Start node (time 0), compute:
–  Earliest Node Time ET for any node is the maximum of the earliest finish
times of all activities immediately preceding activity i: an activity starts at
its tail node’s time: i.e., once all its predecessors have completed
–  Earliest Finish Time EFT= (Earliest Tail Node i Time) + (Duration of
activity ij ).
ET is 0 for nodes with no predecessors, such as the Start node.
Note: The project completion time is the maximum of the Earliest Finish Times
at the Finish node.
In the following, each node represents the completion of an activity. The Earliest
time for the node is written on the the lower right.
Node # |__________________________________________
| ET = max{EFT(=EST+duration) for all predecessors}
18
Earliest Times: Horizon Cable
F, 4
2
6
C, 3
B, 4
Start
D, 2
A, 2 1
2
So the total project time is 19 days.
Finish, 19
E, 10
3
9
Activity
Duration (weeks)
A
2
B
4
C
3
D
2
E
10
F
4
19
Latest Node Times
Step 4: Make a backwards pass through the network, and at each node apply the
Latest Finish Time Rule: LT = Minimum (LT-duration) of the immediate successor nodes
as follows:
(The immediate successors for a node are all nodes that are at the head of an arrow
issuing from the current node.)
Procedure for obtaining latest times for all activities:
•  For the finish node, set LT equal to project completion time.
•  For each activity issuing from a given prior node, calculate
(LT of its finish node) – (duration of the current activity)
•  Apply the latest finish time rule.
•  Repeat until LFT and LST have been obtained for all activities.
Note: The minimum of ST for nodes directly subsequent to the Start Node should
be just the duration of the activity leading to them!
20
Finish Times
Node # | LT
| ET
E.g.: the red 5 is the minimum of the blue 7-1=6, 12-3=9 and
15-10=5
#j
1
Node #i
7 LTj
5 LTi
3
10
#k 12 LTk
#l
15 LTl
21
Example: Horizon Cable
F, 4
2 6
6
C, 3
B, 4
Start
A, 2 1 2
D, 2
Finish, 19
E, 10
3 9
9
2
Note: The only path through the network with zero total slack activities (i.e., no leeway in
start and finish times) is:
A–B–C–E
This is the critical path. There is always at least one critical path . In this example all nodes
are zero slack also.
22
Activity Times and Slack
Each activity has a start node and an end node. In the example
below, there are different earliest times and latest times in each of
nodes i and j relevant to the given activity (i,j):
i
3 LST (i,j), d=4
2 EST
Start Node
LST=latest Start node time
EST=earliest Start node time
j
9 LFT
7 EFT
Final Node
LFT=latest Final node time
EFT=earliest Final node time
Definition of Slack:
The total slack in the activity is 9-(2+4)=3 = LFT – (EST+d)
The free slack in the activity is 7-(2+4)=1 = EFT – (EST+d)
The independent slack in the activity is 7-(3+4)=0 = EFT – (LST+d)
Note that these are in decreasing order: this is always true, as
LFT>EFT and LST>EST.
23
Meaning of the Different Slacks
Total Slack, or Float The extra time that could be made available to an activity,
assuming it has started at its earliest, without delaying the whole project time
Total Slack = LFT – (EST+d)
Free Slack, or Float The extra time that could be made available to an activity,
assuming it has started at its earliest, without preventing beginning at the earliest any
subsequent activity (i.e., not just the final time as with total slack)
Free Slack = EFT – (EST+d)
Independent Slack The extra time that could be made available to an activity,
assuming it has started at its latest, without preventing beginning at the earliest any
subsequent activity
Independent Slack = EFT – (LST+d)
Any path through the network (from Start to Finish) with all zero total slack
activities is a critical path.
24
Effect of Various Slacks in Decision-Making
!   If we use up the total slack in any activity, doing so may modify the critical path
because of its influence on later activities. The coordinator of the whole project is
therefore responsible for the total slack.
!   Free slack does not influence subsequent activities, but it may not be completely
available, unless prior activities are started at their earliest.
!   Independent slack is a property of the activity alone: it does not require that any
other activity be started at its earliest, and it does not affect the available
durations of later activities. So this is the internal slack available to the person(s)
responsible for that activity alone.
25
Example: Horizon Cable
F, 4
2 6
6
C, 3
B, 4
Start
A, 2 1 2
D, 2
Finish, 19
E, 10
3 9
9
2
Activity
A
B
C
D
E
F
Total Slack
0
0
0
1
0
9
Free Slack
0
0
0
1
0
9
Independent Slack
0
0
0
1
0
9
Cr
Cr
Cr
Cr
26
CPM: Crashing a Project
! We use the following notation:
! the normal time to complete an activity (i,j) is Dij , and the normal
cost of performing the activity in that time is cij .
! The activity can be “crashed” by finishing it at a reduced time. The
minimum time for the activity is dij; for that time the (increased)
cost, is Cij.
! In the Critical Path Method (CPM) the following is assumed:
! The relationship between activity (i,j)'s actual crash time, xij, to its
cost is assumed to be linear. We define the overall cost per unit
time reduction, to be change in cost divided by change in time:
Kij = (Cij- cij)/(Dij-dij)
This gives us an equation for the cost yij for any crash time xij :
27
CPM: Crashing a Project
The slope of this line is the negative
(why?) of Kij, so the cost yij for time xij
is given by the slope-intercept
formula:
Lij
yij = - Kij xij + Lij, (i.e., in y = mx + b
form)
Crash cost Cij
where Lij is the (as yet unknown) yintercept of this line.
actual cost yij
Normal cost cij
dij
Actual time: xij
Dij
28
CPM: Crashing a Project
Thus the total cost is the sum of the costs of crashing along all activities:
Cost = Σ yij = - ΣKij xij + ΣLij)
A final assumption is that
! The critical path does not change from that obtained by using the normal times,
when some or all of the activities are crashed within their ranges.
So we want to minimize the total cost, i.e., we want to minimize – ΣKij xij + ΣLij. But
this is the same as maximizing its negative:
Maximize ΣKij xij - ΣLij
Finally, the sum of the y-intercepts, ΣLij , is a constant which does not affect the
optimization process, so we can ignore this piece for now and all we actually need to
do is to
29
CPM Theory Continued
Maximize Z = ΣKij xij
This looks like it is becoming an LP problem and indeed it is: the constraints on this
maximization are straightforward:
Along the critical path, we need to ensure that no activity starts before any prior
activity has finished. So we introduce (variable) node times ti for each node #i on the
critical path. Then it is clear that node j cannot happen until at least activity (i,j) is
done:
tj ≥ ti + xij or ti ≤ tj - xij
(where by ≤ we mean “less than or equal to” etc.) Also, each of the ti is non-negative,
and we need to ensure that the project is completed by the pre-determined (crashed)
project time.
Node #i
ti
xij
Node #j
tj
30
The CPM LP
Finally, each activity time has to be greater than or equal to the minimum, completely
crashed time given. This is always at least zero, so we do not need non-negativity
constraints! So we have formulated completely the CPM problem of time-cost
tradeoffs:
In order to crash the project so as to achieve the stated project time, with
minimal increase in cost, solve the following LP:
The CPM Linear Programming Problem
Variables: Activity times xij, Node completion times ti, (including final node tFinal Node )
Maximize Z = ΣKij xij
Subject to:
ti ≤ tj - xij for all critical activities (i,j), and
tFinal Node ≤ Desired Project Time
xij ≥ dij, the minimum time allowed for activity (i,j)
xij ≤ Dij, the maximum time allowed for activity (i,j)
31
Example: Auditing a Corporation
! 
When an accounting firm audits a corporation, the first phase of the audit involves obtaining "knowledge 0f the business".
This requires the activities in below. Draw the activity network, determine the critical path, and find the total float and free
float for each activity.
Activity
A
B
C
D
E
F
G
Description
Determining
terms of
engagement
Appraisal of
auditability
risk
Identification
of types of
transactions
and possible
errors
Systems
description
Verification of
systems
description
Evaluation of
internal
controls
Design of
Audit
approach
Immediate
Predecessors
-
A
A
C
D
B,E
F
Duration
(days)
3
6
14
8
4
8
9
32
Example for CPM: The Audit
The network is:
5
2
3
3
A, 3
B, 6
E, 4
17
6
37
37
17
Start
D, 8
F, 8
29
C, 14
3
29
4
25
G, 9
25
Critical Path: ACDEFG. Nonzero slack only for B: all three slacks are 20.
Now let’s see what happens when we have to crash the project:
Finish, 46
33
Example: Crashing the Audit
Assume the auditing project must be completed in 30 days. The maximum
allowable reduction in duration, as well as the cost per day of such a
reduction, is appended below. Formulate an LP
that can be used to minimize the cost of meeting the projected deadline.
Activity
A
B
C
D
E
F
G
Normal Duration
(days): Dij
3
6
14
8
4
8
9
Cost per day of
reducing duration: Kij
$100
$80
$60
$70
$30
$20
$50
Maximum allowable
reduction in duration
3
4
5
2
4
4
4
(Minimum time: dij
0
2
9
6
0
4
5)
34
Example for CPM: The Audit
!   Subtracting the third row above from the second, we get the minimum times, dij. The cost
per day of reduction, i.e., the Kij , are given in the second column above. Finally, the critical
path is ACDEFG, so the LP for completing the project by the deadline of 30 days is:
!   Maximize Z = 100 xA + 80 xB + 60 xC + 70 xD + 30 xE + 20 xF + 50 xG
Subject to:
network time constraints along the critical path:
xA – t2 ≤ 0; xB + t2 – t5 ≤ 0; xC + t2 – t3 ≤ 0;
xD + t3 – t4 ≤ 0; xE + t4 – t5 ≤ 0; xF + t5 – t6 ≤ 0; xG + t6 – t7 ≤ 0;
the total time constraint:
t7 ≤ 30;
minimum time constraints:
xA ≥ 0; xB ≥ 2; xC ≥ 9; xD ≥ 6; xE ≥ 0; xF ≥ 4; xG ≥ 5;
and maximum time constraints:
xA ≤ 3; xB ≤ 6; xC ≤ 14; xD ≤ 8; xE ≤ 4; xF ≤ 8; xG ≤ 9.
Note: It is important to include all the activities in Z, since this is a measure of total cost. Similarly, we must include
all activities in the minimum–time constraints. Only the network constraints are restricted to the critical path.
35
Example for CPM: The Audit – Solution of the LP
Using MAPLE, the solution is: In order to complete the project in 30 days, run each
activity according to the red numbers:
5
A,
3, 3
2
3
3
3
B, 6, 6
C, 14,
10
3 17
13
29
21
E, 4, 0
6
25
17
Start
D, 8,
8
F, 8, 4
29
4
21
25
37
37
G, 9,
5
25
With Z = $2,270.
We have also entered in blue the completion times for each node.
Finish, 46
30
36
Calculating the Increased Cost
The quantity to be minimized, is the total cost, which is given by the formula on slide 31:
- ΣKij xij + ΣLij = Σ Lij – Z. Thus, if we can compute Σ Lij for all activities in the network, we will
be able to find the minimum total cost.
! For the line on slide 30, we know that the equation of the cost-duration line is
yij = - Kij xij + Lij, and we know that the normal time-and-cost point (Dij , cij) is on this line.
Plugging that in and solving for Lij, we see that
Lij = cij + Kij Dij
Thus the total cost is = Σ( cij + Kij Dij ) – Z
What does this mean?
37
Formula for Total Increased Cost; Meaning of Z
!
Total cost = Σ( cij + Kij Dij ) – Z = Σ cij + Σ Kij Dij – Z
!
!
= Normal-Time Cost + (Cost/unit reduction)X(Normal Time) - Z
= Normal-Time Cost + Virtual Cost of reducing all times to zero - Z
!
!
Thus the total increased cost is Σ Kij Dij – Z , i.e., the
Virtual Cost of reducing all times to zero - Z
!   So Z turns out to be the correction we have to apply because some or all of the actual
computed times are bigger than zero.
!   To find the increased cost in any actual example, introduce a new row which calculates the
quantity Kij Dij for each activity, add and finally subtract Z.
!   Another way to do this is to calculate each actual reduction in time, and multiply by the
corresponding cost-per-unit reduction, and then add.
38
Example for CPM: The Audit – Solution of the LP
!   For this example, we had found Z = $3,390. To the table of normal times, introduce a new
row for the product (Cost/unit reduction)X(Normal Time) and add along that row:
Activity
A
B
C
D
E
F
G
Normal Duration
(days): Dij
3
6
14
8
4
8
9
Cost per day of
reducing duration: Kij
$100
$80
$60
$70
$30
$20
$50
Maximum allowable
reduction in duration
3
4
5
2
4
4
4
(Minimum time: dij
0
2
9
6
0
4
5)
Row 1 X Row 2
300
480
840
560
120
160
450
So the first term is the sum along this last row: $2,910. This is what it would have cost had we
crashed every activity to zero (of course, this is disallowed by the minimum time
requirements!). Subtracting Z= $2,270, we get that the
Total minimal increased cost for the audit is $640.
39
Program Evaluation Research Technique
!   In practice, activity times are somewhat random.
!   A good estimate for the distribution of an activity time can be made on the basis of
past experience.
!   In some instances there is no past experience but a best guess at the distribution
of activity times can still be made.
!   It turns out that, in general, the randomness of activity times is well approximated
by the so-called “beta” distribution. This is a distribution that is completely
determined by three parameters: the shortest, or optimistic time a, the longest, or
pessimistic time b and the mode, or most likely time m.
!   In the following we will show how knowing these three numbers for each activity
will allow an estimate of the likelihood of meeting a specific completion date.
!   This is therefore also called the “three estimate approach.”
40
The Beta Distribution
1
0.8
y
0.6
0.4
0.2
0
0
1
2
3
4
x
5
6
7
8
Here is a typical Beta distribution, with a = 1, b = 7 and m approximately 3. (Why is
the shape of the beta distribution (i.e., skewed to the right) reasonable for activity
times?)
!   Fact: For any beta distribution with parameters a, m and b, the mean is:
t = (a + 4m + b)/6
And the variance (i.e., square of the standard deviation) is
V = (b-a)2/36
41
Sums of Independent Random Variables
!   Here’s another fact from probability theory:
Sums of Independent Random Variables
!   If X1, X2…Xn are independent random variables with means µ1, µ2…µn and
variances V1, V2,…,Vn respectively, then for the sum random variable
X = X1 + X2 +…+ Xn,
the mean is µ1 + µ2 +…+ µn and the variance is V1 + V2 + …+ Vn:
Given independence, the means and variances add
Note: The standard deviations do not add to give the total standard deviation! We have to first add their
squares (.e., the variances), and then take the square root, to find the “total” standard deviation.
So:
The total project time is a random variable, that has a mean and a variance that
are both sums of the same along the critical path
42
Central Limit Theorem
!   Here’s another fact from probability theory:
Central Limit Theorem
The sum of a (sufficiently large) number of independent random variables is
approximately normally distributed
!   We are assuming that the critical path does not change in the range (a to b) of
activity times given by the parameters for the set of activities
!   We find the critical path by using the modal times m for the activities.
!   Thus the project completion time random variable will have a normal distribution
with mean equal to the sum of the means along the critical path and variance
equal to the sum of the variances along the critical path.
43
PERT Method
1 
2 
3 
4 
Draw the activity network, as in CPM
Work out the critical path, using the modal times m for each activity, as in CPM
Find and then add the means only along the critical path, using the formula
µ = Σ (a + 4m + b)/6
Find and then add the Variances only along the critical path, using the formula
V = Σ (b-a)2/36
5 
6 
Find σ = the square root of V
The total time X along the critical path is now a normally distributed random variable with
mean µ (the expected project time) and standard deviation σ.
7 
Find the probability that the random total project time variable X is less than, or equal to,
any given specified completion time T, i.e., P(X <= T) by converting to Z-scores and using
the standard normal distribution tables.
44
Example: #7 on the Handout
4 13
2 10
13
7 22
6
22
5 20
1 0
14
0
9 37
3 4
8 28
4
19
37
6 25
16
45
Example: #7 on the Handout
!   The above is the full network with times computed using the m’s from the table
given in the problem, giving a total time of 37. Here it is clear that the critical path
is
1-3-4-7-9
and that the floats (slacks) are:
Activity
(1,2)
(1,3)
(2,4)
(3,4)
(3,5)
(3,6)
4,7)
(5,7)
(6,8)
(7,9)
(8,9)
Total
Float
4
0
4
0
6
9
0
6
9
0
9
Free
Float
0
0
4
0
0
0
0
6
0
0
9
Inde- 0
pendent
Float
0
0
0
0
0
0
0
0
0
0
46
Example: #7 on the Handout
!   We next compute the means and variances of each of the activities along the
critical path 1-3-4-7-9:
Activity
(1,3)
(3,4)
(4,7)
(7,9)
Sum
Mean
6
9
8.83
15
38.83
Variance
1
1
1.36
2.78
6.14
!   Notice that the mean along the critical path is slightly greater than the length
computed using the m’s. This is reasonable for a distribution that is skewed to the
right.
!   Thus the project time is N(38.83, 2.48) (the standard deviation is the square root
of the variance of 6.14).
47
Example: #7 on the Handout
!   Thus the probability that this normal random variable is less than or equal to 40 is
given by the probability that a standard normal random variable is les than or
equal to
!   Z = (40 - 38.83)/2.48 = 0.47
!   From the tables, this is 0.6808:
!   Answer: The probability that the project will be completed in 40 days is 68%
48